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2 years ago

What is the total number of moles of H2SO4 needed to prepare 5.0 liters of a 2.0 M solution of H2SO4

Chemistry

2 answers:

Lerok [7]2 years ago

5 0

Answer:

Explanation:

forsale [732]2 years ago

3 0

The molarity of H2SO4 is the number of moles in 1 L of solution.
The molarity is 2.0 mol/L
This means that there should be 2 moles in a 1 L solution to make up this molarity.
In this case we need to make up a 5 L solutions with that molarity. Then the amount of moles required are - 2 mol/L x 5 L = 10 mol

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The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma

IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

$A=6R^{2}\sqrt{3}$

Here, R is radius and is related to a as follows:

$R=\frac{a}{2}$

Putting the value in expression for area,

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

The value of a is 0.4961 nm

Since, $1 nm=10^{-7}cm$

Thus, $0.4961 nm=4.961\times 10^{-8} cm$

Putting the value,

$Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}$

Now, volume can be calculated as follows:

$V=Area\times c$

The value of c is 1.360 nm or $1.360\times 10^{-7} cm$

Putting the value,

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

now, number of atom in unit cell can be calculated by using the following formula:

$n=\frac{\rho N_{A}V_{c}}{A}$

Here, A is atomic mass of $Cr_{2}O_{3}$ is 151.99 g/mol.

Putting all the values,

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Thus, there will be 18 $Cr_{2}O_{3}$ units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of $Cr^{3+}$ and $O^{2-}$ is 62 pm and 140 pm respectively.

Converting them into cm:

$1 pm=10^{-10}cm$

Thus,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

and,

$r_{O^{2-}}=1.4\times 10^{-8}cm$

Volume of sphere will be sum of volume of total number of cations and anions thus,

$V_{S}=V_{Cr^{3+}}+V_{O^{2-}}$

Since, volume of sphere is $V=\frac{4}{3}\pi r^{3}$ ,

$V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )$

Putting the values,

$V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}$

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

$packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758$

Thus, atomic packing factor is 0.758.

6 0

2 years ago

Read 2 more answers

A 0.1064 g sample of a pesticide was decomposed by the action of sodium biphenyl. The liberated Cl- was extracted with water and

Ilya [14]

Answer:

Percentage of an aldrin in the sample is 44.41%.

Explanation:

$Cl^-+AgNO_3\rightarrow AgCl+NO_{3}^-$

Molarity of the silver nitrate solution = 0.03337 M

Volume of the silver nitrate = 23.28 mL = 0.02328 L

Moles of silver nitrate = n

$0.03337 M=\frac{n}{0.02328 L}$

n = $0.03337 M\times 0.02328 L=0.0007768 mol$

According to reaction 1 mole of silver nitrate recats with 1 moles of chloride ions.

Then 0.0007768 moles of silver nitrate will react with:

$\frac{1}{1}\times 0.0007768 mol=0.0007768 mol$ chloride ions.

In one mole of aldrin there are 6 moles of chloride ions.

Then moles of aldrin containing 0.0007768 moles chloride ions are:

$\frac{0.0007768 mol}{6}=0.0001295 mol$

Moles of aldrin present in the sample = 0.0001295 mol

Mass of 0.0001295 moles of aldrin present in the sample :

0.0001295 mol × 364.92 g/mol =0.04726 g

Percentage of an aldrin in the sample:

$\frac{0.04726 g}{0.1064 g}\times 100=44.41\%$

8 0

2 years ago

A 23.2 g sample of an organic compound containing carbon, hydrogen and oxygen was burned in excess oxygen and yielded 52.8 g of

allochka39001 [22]

Answer:

The answer to your question is C₃H₆O

Explanation:

Data

mass of sample = 23.2 g

mass of carbon dioxide = 52.8 g

mass of water = 21.6 g

empirical formula = ?

Process

1.- Calculate the mass and moles of carbon

44 g of CO₂ --------------- 12 g of C

52.8 g --------------- x

x = (52.8 x 12)/44

x = 633.6/44

x = 14.4 g of C

12 g of C ------------------ 1 mol

14.4 g of C --------------- x

x = (14.4 x 1)/(12)

x = 1.2 moles of C

2.- Calculate the grams and moles of Hydrogen

18 g of H₂O --------------- 2 g of H

21.6 g of H₂O ------------- x

x = (21.6 x 2) / 18

x = 2.4 g of H

1 g of H -------------------- 1 mol of H

2.4 g of H ----------------- x

x = (2.4 x 1)/1

x = 2.4 moles of H

3.- Calculate the grams and moles of Oxygen

Mass of Oxygen = 23.2 - 14.4 - 2.4

= 6.4 g

16 g of O ---------------- 1 mol

6.4 g of O -------------- x

x = (6.4 x 1)/16

x = 0.4 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon = 1.2 / 0.4 = 3

Hydrogen = 2.4/ 0.4 = 6

Oxygen = 0.4 / 0.4 = 1

5.- Write the empirical formula

C₃H₆O

8 0

2 years ago

What is the concentration of a solution made with 0.150 moles of KOH in 400.0 mL of solution?

otez555 [7]

Answer:

$concentration = \frac{0.15}{0.4}=0.375 mol/L$

Explanation:

Concentration: i is defined as the mole per litre.

$concentration = \frac{mole}{volume in L}$

mole=0.15

volume=400 ml=0.4 litre

$concentration = \frac{0.15}{0.4}=0.375 mol/L$

6 0

2 years ago

What is the molarity of a solution that contains 2.35 g of nh3 in 0.0500 l of solution?

tankabanditka [31]

The molarity is the number of moles in 1 L of the solution.
The mass of NH₃ given - 2.35 g
Molar mass of NH₃ - 17 g/mol
The number of NH₃ moles in 2.35 g - 2.35 g / 17 g/mol = 0.138 mol
The number of moles in 0.05 L solution - 0.138 mol
Therefore number of moles in 1 L - 0.138 mol / 0.05 L x 1L = 2.76 mol
Therefore molarity of NH₃ - 2.76 M

3 0

2 years ago

Read 2 more answers